Question

Help with the following discrete math question?!

. Prove that the following four statements are equivalent:
(a) n^2 is odd.
(b) 1 − n is even.
(c) n^3 is odd.
(d) n^2 + 1 is even.

Answer 1

so first you have to see that this is not true for all n.

Answer 2

By this question, they are indeed asking if all statements hold the same truth value. To be equivalent here, they must either all be false or all be true.

Answer 3

k lets look at, what does it mean for a number to be odd?

Answer 4

well, thing is my teacher always used the definition of n=2k being even. That was always how he s tarted and how my brain sees it. Im having a hard time starting this problem cause of that as n seems to be odd because n^2 is odd

Answer 5

yeah, for a number to be odd, it has to be of the form 2i+1 for an integer i. so for n^2 to be odd it has to be of the form (2i+1)^2

Answer 6

= \[4i ^{2}+4i+1\]

Answer 7

and then find its root Im guessing and it should give me (2i+1) or something right? showibng that its odd

Answer 8

you can just make the assumption that a is true, and then show that if a is true then b,c,&d must also be true

Answer 9

so I assume n^2 is odd, that means that it is of the form 4k^2+4k+1, you should see that that number is always odd,

Answer 10

ohhh I see if n^2 is odd, then n is odd therefore n-1 must be true ( if its hold, so I have to show it) and then n^3 must be odd and so on?

Answer 11

yeah, ill show you the easiest one, d. if n^2 is odd, it = 4k^2+4k+1. so n^2+1 is odd,
n^2+1 = (4k^2+4k+1)+1 =(4k^2+4k+2)=2(2k^2+2k+1)

Answer 12

2(2k^2+2k+1) has to be even, because it is divisible by 2

Answer 13

mhm I can see what you're saying by being divisible by 2 , however shouldn't it be proven by using the initial definition?

Answer 14

it was, it was built upon the assumption that a was true, ie n = 2k+1

Answer 15

alright thanks a lot, very helpful!